Answer:
∠ABC = 76°
BC = 20.1
CA = 28.0
Explanation:
Solving the triangle means finding all unknown angles and sides of the triangle.
(i) Two of the angles (∠BCA = 60° and ∠CAB = 44°) are given. To find the third angle (∠ABC), use one of the theorems stating that the sum of angles of a triangle is equal to 180°.
Therefore, the sum of angles of the triangle ABC is 180°. i.e
∠ABC + ∠BCA + ∠CAB = 180°
=> ∠ABC + 60° + 44° = 180°
=> ∠ABC + 104° = 180°
=> ∠ABC = 180° - 104°
=> ∠ABC = 76°
(ii) One side (BA) of the triangle is given. To get the other sides, we use the sine rule as follows;
=>
![(sin60)/(25) = (sin44)/(BC) = (sin76)/(CA)](https://img.qammunity.org/2022/formulas/mathematics/high-school/tc1yj5qolmi6eandnk2ixf6d35xeeg07af.png)
=>
![(sinBCA)/(BA) = (sinCAB)/(BC) = (sinABC)/(CA)](https://img.qammunity.org/2022/formulas/mathematics/high-school/xypnh40v0tbncxbbxqvoawhr1x0wdwka8b.png)
Substitute the necessary values
---------------------(ii)
(a) To get side BC, use the first two terms of equation (ii)
![(sin60)/(25) = (sin44)/(BC)](https://img.qammunity.org/2022/formulas/mathematics/high-school/5a0og0bs5tleh066zzk1jd0fzbkzradhus.png)
Cross multiply
BC x sin 60 = 25 x sin 44
BC x 0.8660 = 25 x 0.6947
0.8660 x BC = 17.3675
BC =
![(17.3675)/(0.8660)](https://img.qammunity.org/2022/formulas/mathematics/high-school/89ysuju2davsb0qcnh5pd21dqqcd7jtv5i.png)
BC = 20.05
=> BC = 20.1 to the nearest tenth
(b) To get CA, use any two terms of equation (ii). Using the first and third terms, we have;
![(sin60)/(25) = (sin76)/(CA)](https://img.qammunity.org/2022/formulas/mathematics/high-school/eneijya9p5y8qfunehuqtsl3zmbw72vcgr.png)
Cross multiply
CA x sin 60 = 25 x sin 76
CA x 0.8660 = 25 x 0.9703
0.8660 x CA = 24.2575
CA =
![(24.2575)/(0.8660)](https://img.qammunity.org/2022/formulas/mathematics/high-school/qfyevj66pd0t1t3u0y759jmz2hz0rvb7mz.png)
CA = 28.01
=> CA = 28.0 to the nearest tenth